~ 2 min read

Getting 1/3 Probability by Tossing Coins -- Solution


How can we choose one out of 3 christmass presents with equal probabability?  Or in other words how to obtain 1/3 probability by using one unbiased coin ?


1. The coin is UNbiased: toss the coin twice. Let TH, HT and HH correspond to each of the presents. In case of HH - repeat from the beginning.

2. The coin is biased, then we notice that TH and HT would occur with equal probability but this will only give us 2 possible outcomes with equal probability and we need 3.  Then we would have to use 4 tosses which will give us 16 possible combinations. The question is which ones we should choose to represent the presents (or outcomes). The only condition is that the number of H and T should be the same to ensure equal probability for each outcome and thus selecting each present with the same probability. One option would be to assign HTHT, THTH and HTTH to the three choices, with other  13 possible 4-toss outcomes being rejected.


So the obvious problem is that we need some way to extrapolate tosses of a single coin to (possibly) arbitrary number of outcomes with equal probability to obtain each outcome.  So obviously we will need several tosses to match the number of outcomes we need. However we need to pay careful attention to the fact that the toss sequences we are using have the same combined probability such as HHTT and HTHT